Answer :
To predict the products of the combustion reaction and balance the chemical equation, follow these steps:
1. Identify the Reactants:
The reactants are:
- [tex]\((CH_3)_3CH_4\)[/tex] (a hydrocarbon in gaseous form)
- [tex]\(O_2\)[/tex] (oxygen gas)
2. Predict the Products of Combustion:
Combustion of a hydrocarbon typically produces:
- Carbon dioxide ([tex]\(CO_2\)[/tex])
- Water ([tex]\(H_2O\)[/tex])
3. Write the Unbalanced Equation:
Write the unbalanced chemical equation for the reaction:
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow CO_2 + H_2O $[/tex]
4. Balance the Chemical Equation:
To balance the chemical equation, ensure that the number of atoms of each element on the reactant side equals the number on the product side. Let's break down the balancing process:
i. Identify the number of each type of atom in the reactants and products:
- In [tex]\((CH_3)_3CH_4\)[/tex], there is a total of 10 Carbon atoms and 22 Hydrogen atoms.
- For [tex]\(O_2\)[/tex], initially, there are 2 Oxygen atoms.
- On the product side, we need to balance Carbon, Hydrogen, and Oxygen atoms appropriately.
ii. Balance Carbon atoms:
To balance the carbons, we need 10 [tex]\(CO_2\)[/tex] molecules because there are 10 Carbon atoms in the reactant hydrocarbon.
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow 10CO_2 + H_2O $[/tex]
iii. Balance Hydrogen atoms:
To balance the 22 Hydrogen atoms in the reactant hydrocarbon, we need 11 [tex]\(H_2O\)[/tex] molecules because each molecule of [tex]\(H_2O\)[/tex] contains 2 Hydrogen atoms.
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow 10CO_2 + 11H_2O $[/tex]
iv. Balance Oxygen atoms:
On the product side, each [tex]\(CO_2\)[/tex] molecule has 2 Oxygen atoms and each [tex]\(H_2O\)[/tex] molecule has 1 Oxygen atom:
- From [tex]\(10CO_2\)[/tex]: [tex]\(10 \times 2 = 20\)[/tex] Oxygen atoms
- From [tex]\(11H_2O\)[/tex]: [tex]\(11 \times 1 = 11\)[/tex] Oxygen atoms
- Total Oxygen atoms on the product side: [tex]\(20 + 11 = 31\)[/tex]
On the reactant side, initially [tex]\(O_2\)[/tex] must be adjusted to have 31 Oxygen atoms:
[tex]$ \frac{31}{2} O_2 \rightarrow 15.5 O_2 $[/tex]
To remove the fraction, multiply the entire equation by 2:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
So, the balanced chemical equation is:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
Summary:
The products of the combustion reaction of [tex]\((CH_3)_3CH_4\)[/tex] in the presence of [tex]\(O_2\)[/tex] are carbon dioxide ([tex]\(CO_2\)[/tex]) and water ([tex]\(H_2O\)[/tex]), with the balanced equation being:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
1. Identify the Reactants:
The reactants are:
- [tex]\((CH_3)_3CH_4\)[/tex] (a hydrocarbon in gaseous form)
- [tex]\(O_2\)[/tex] (oxygen gas)
2. Predict the Products of Combustion:
Combustion of a hydrocarbon typically produces:
- Carbon dioxide ([tex]\(CO_2\)[/tex])
- Water ([tex]\(H_2O\)[/tex])
3. Write the Unbalanced Equation:
Write the unbalanced chemical equation for the reaction:
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow CO_2 + H_2O $[/tex]
4. Balance the Chemical Equation:
To balance the chemical equation, ensure that the number of atoms of each element on the reactant side equals the number on the product side. Let's break down the balancing process:
i. Identify the number of each type of atom in the reactants and products:
- In [tex]\((CH_3)_3CH_4\)[/tex], there is a total of 10 Carbon atoms and 22 Hydrogen atoms.
- For [tex]\(O_2\)[/tex], initially, there are 2 Oxygen atoms.
- On the product side, we need to balance Carbon, Hydrogen, and Oxygen atoms appropriately.
ii. Balance Carbon atoms:
To balance the carbons, we need 10 [tex]\(CO_2\)[/tex] molecules because there are 10 Carbon atoms in the reactant hydrocarbon.
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow 10CO_2 + H_2O $[/tex]
iii. Balance Hydrogen atoms:
To balance the 22 Hydrogen atoms in the reactant hydrocarbon, we need 11 [tex]\(H_2O\)[/tex] molecules because each molecule of [tex]\(H_2O\)[/tex] contains 2 Hydrogen atoms.
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow 10CO_2 + 11H_2O $[/tex]
iv. Balance Oxygen atoms:
On the product side, each [tex]\(CO_2\)[/tex] molecule has 2 Oxygen atoms and each [tex]\(H_2O\)[/tex] molecule has 1 Oxygen atom:
- From [tex]\(10CO_2\)[/tex]: [tex]\(10 \times 2 = 20\)[/tex] Oxygen atoms
- From [tex]\(11H_2O\)[/tex]: [tex]\(11 \times 1 = 11\)[/tex] Oxygen atoms
- Total Oxygen atoms on the product side: [tex]\(20 + 11 = 31\)[/tex]
On the reactant side, initially [tex]\(O_2\)[/tex] must be adjusted to have 31 Oxygen atoms:
[tex]$ \frac{31}{2} O_2 \rightarrow 15.5 O_2 $[/tex]
To remove the fraction, multiply the entire equation by 2:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
So, the balanced chemical equation is:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
Summary:
The products of the combustion reaction of [tex]\((CH_3)_3CH_4\)[/tex] in the presence of [tex]\(O_2\)[/tex] are carbon dioxide ([tex]\(CO_2\)[/tex]) and water ([tex]\(H_2O\)[/tex]), with the balanced equation being:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
